TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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A generating family on the codomain suffices to test measurability

Statement

Let $(X, A)$ and $(Y, B)$ be measurable spaces, let $G \subseteq P (Y)$ generate $B$ , and let $f : X \to Y$ be a function. If

$f^{- 1} (G) \in A for every G \in G,$

then $f$ is measurable as a map $(X, A) \to (Y, B)$ .

Facts & Assumptions

Given: Measurable spaces $(X, A)$ and $(Y, B)$ , a function $f : X \to Y$ , and a generating family $G$ for $B$ .

[L1]

A function is measurable exactly when the preimage of every measurable set in the codomain is measurable in the domain. (A measurable function between measurable spaces)

Proof

technique · direct

1.1

Let

L1algebra

$C : = {B \subseteq Y : f^{- 1} (B) \in A} .$

Preimages preserve complements and countable unions, so $C$ is a sigma-algebra on $Y$ . [L1, algebra]

2.1

By hypothesis, every member of $G$ lies in $C$ . Since [given, step 1.1] $B$ is the sigma-algebra generated by $G$ , one has $B \subseteq C$ .

givenstep 1.1

3.1

Therefore $f^{- 1} (B) \in A$ for every $B \in B$ , and [step 2.1, L1] [L1] says exactly that $f$ is measurable.

step 2.1L1∎

Depends on

A measurable function between measurable spaces

Used by

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Measure, Integration and Real Analysis, Section 2B (standard reference, not scraped)
John K. Hunter, Measure Theory, Proposition 3.2 (standard reference, not scraped)